How to add and extend a new row in Python list?
I have to reduce the density of an array, by using a for loop that traverses by steps of 100 and copy the value of my original array into a new array:
soundDataHere is a [7][22000] dim array, and I want cleanSoundData to be a [7][220] dim array
def reduceDensity(soundDataHere):
for i in range(numberOfFiles):
for j in range(0, soundDataHere[i].size-1, 100):
cleanSoundData.extend(soundDataHere[i][j])
I keep dont know how to use the append and extend function in a foor loop to recreate a new less dense array.
example: [[1,2,3,4,5],[6,7,8,9,10]] with a step = 2
should return [[1,3,5],[6,8,10]] in my new cleanSoundData array
but is only extending it like [1,3,5,6,8,10]
python arrays
edited Dec 29 '18 at 16:02
Sergey Pugach
1,8021412
asked Dec 29 '18 at 8:54
Sharan NarasimhanSharan Narasimhan
218
add a comment |
I have to reduce the density of an array, by using a for loop that traverses by steps of 100 and copy the value of my original array into a new array:
soundDataHere is a [7][22000] dim array, and I want cleanSoundData to be a [7][220] dim array
def reduceDensity(soundDataHere):
for i in range(numberOfFiles):
for j in range(0, soundDataHere[i].size-1, 100):
cleanSoundData.extend(soundDataHere[i][j])
I keep dont know how to use the append and extend function in a foor loop to recreate a new less dense array.
example: [[1,2,3,4,5],[6,7,8,9,10]] with a step = 2
should return [[1,3,5],[6,8,10]] in my new cleanSoundData array
but is only extending it like [1,3,5,6,8,10]
python arrays
edited Dec 29 '18 at 16:02
Sergey Pugach
1,8021412
asked Dec 29 '18 at 8:54
Sharan NarasimhanSharan Narasimhan
218
Is using the numpy module an option for you?
– Jason Baumgartner
Dec 29 '18 at 8:58
add a comment |
I have to reduce the density of an array, by using a for loop that traverses by steps of 100 and copy the value of my original array into a new array:
soundDataHere is a [7][22000] dim array, and I want cleanSoundData to be a [7][220] dim array
def reduceDensity(soundDataHere):
for i in range(numberOfFiles):
for j in range(0, soundDataHere[i].size-1, 100):
cleanSoundData.extend(soundDataHere[i][j])
I keep dont know how to use the append and extend function in a foor loop to recreate a new less dense array.
example: [[1,2,3,4,5],[6,7,8,9,10]] with a step = 2
should return [[1,3,5],[6,8,10]] in my new cleanSoundData array
but is only extending it like [1,3,5,6,8,10]
python arrays
edited Dec 29 '18 at 16:02
Sergey Pugach
1,8021412
asked Dec 29 '18 at 8:54
Sharan NarasimhanSharan Narasimhan
218
I have to reduce the density of an array, by using a for loop that traverses by steps of 100 and copy the value of my original array into a new array:
soundDataHere is a [7][22000] dim array, and I want cleanSoundData to be a [7][220] dim array
def reduceDensity(soundDataHere):
for i in range(numberOfFiles):
for j in range(0, soundDataHere[i].size-1, 100):
cleanSoundData.extend(soundDataHere[i][j])
I keep dont know how to use the append and extend function in a foor loop to recreate a new less dense array.
example: [[1,2,3,4,5],[6,7,8,9,10]] with a step = 2
should return [[1,3,5],[6,8,10]] in my new cleanSoundData array
but is only extending it like [1,3,5,6,8,10]
python arrays
python arrays
edited Dec 29 '18 at 16:02
Sergey Pugach
1,8021412
asked Dec 29 '18 at 8:54
Sharan NarasimhanSharan Narasimhan
218
edited Dec 29 '18 at 16:02
Sergey Pugach
1,8021412
asked Dec 29 '18 at 8:54
Sharan NarasimhanSharan Narasimhan
218
edited Dec 29 '18 at 16:02
Sergey Pugach
1,8021412
edited Dec 29 '18 at 16:02
Sergey Pugach
1,8021412
edited Dec 29 '18 at 16:02
Sergey Pugach
1,8021412
1,8021412
asked Dec 29 '18 at 8:54
Sharan NarasimhanSharan Narasimhan
218
asked Dec 29 '18 at 8:54
Sharan NarasimhanSharan Narasimhan
218
asked Dec 29 '18 at 8:54
Sharan NarasimhanSharan Narasimhan
218
218
Is using the numpy module an option for you?
– Jason Baumgartner
Dec 29 '18 at 8:58
add a comment |
Is using the numpy module an option for you?
– Jason Baumgartner
Dec 29 '18 at 8:58
Is using the numpy module an option for you?
– Jason Baumgartner
Dec 29 '18 at 8:58
Is using the numpy module an option for you?
– Jason Baumgartner
Dec 29 '18 at 8:58
add a comment |
3 Answers
3
active
oldest
votes
Maybe you should try creating two different temporary list objects where you 'extend' the required elements into each one of them(that is the right and the left sublist) just after the for loop. Then 'append' those two lists instead of extending in the cleanSoundData.
answered Dec 29 '18 at 9:08
Aditya GoyalAditya Goyal
11
thats possible but more memory hungry and slower, if theres another better method to do it that would be preferred.
– Sharan Narasimhan
Dec 29 '18 at 9:12
add a comment |
Using Numpy and your example:
import numpy as np
l = [1,3,5,6,8,10]
l2 = np.reshape(l,[2,-1])
>>> l2
array([[ 1, 3, 5],[ 6, 8, 10]])
It looks like you're working with sound data so I would highly recommend using the numpy module as vectorizing array operations will be far faster than using a for loop with Python objects (up to 100x faster in some cases).
The Numpy module was designed specifically for these type of use cases so I would encourage you to learn how to use it.
answered Dec 29 '18 at 9:03
Jason BaumgartnerJason Baumgartner
292110
Yes okay I will consider using numpy, however its tough to use reshape all the time because all sound samples do not have the same size. But thanks I can probably work around this problem with a little bit more code
– Sharan Narasimhan
Dec 29 '18 at 9:16
np.reshape(l,[2,-1]) -- the -1 here means to auto-adjust to the size so as long as the array is even sized, it will work. If you could give more info on the structure of the data, I might be able to give you some better ideas.
– Jason Baumgartner
Dec 29 '18 at 9:21
add a comment |
Assuming that the array soundData is a 7 X 22000 holds your data. So creating a new array cleanSoundData of size 7 x 220 can be done as follows. Even it is more generalized whether its 2 x 1000 or 1000 x 50000.
cleanSoundData =
for i in range(len(soundData)):
cleanSoundData.append() # adding new row
for j in range(0, len(soundData[i]), 100):
cleanSoundData[i].append(soundData[i][j]) # adding data to the row
Hope this will work for you.
answered Dec 29 '18 at 9:26
Anidhya BhatnagarAnidhya Bhatnagar
46329
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Maybe you should try creating two different temporary list objects where you 'extend' the required elements into each one of them(that is the right and the left sublist) just after the for loop. Then 'append' those two lists instead of extending in the cleanSoundData.
answered Dec 29 '18 at 9:08
Aditya GoyalAditya Goyal
11
thats possible but more memory hungry and slower, if theres another better method to do it that would be preferred.
– Sharan Narasimhan
Dec 29 '18 at 9:12
add a comment |
Maybe you should try creating two different temporary list objects where you 'extend' the required elements into each one of them(that is the right and the left sublist) just after the for loop. Then 'append' those two lists instead of extending in the cleanSoundData.
answered Dec 29 '18 at 9:08
Aditya GoyalAditya Goyal
11
thats possible but more memory hungry and slower, if theres another better method to do it that would be preferred.
– Sharan Narasimhan
Dec 29 '18 at 9:12
add a comment |
Maybe you should try creating two different temporary list objects where you 'extend' the required elements into each one of them(that is the right and the left sublist) just after the for loop. Then 'append' those two lists instead of extending in the cleanSoundData.
answered Dec 29 '18 at 9:08
Aditya GoyalAditya Goyal
11
Maybe you should try creating two different temporary list objects where you 'extend' the required elements into each one of them(that is the right and the left sublist) just after the for loop. Then 'append' those two lists instead of extending in the cleanSoundData.
answered Dec 29 '18 at 9:08
Aditya GoyalAditya Goyal
11
answered Dec 29 '18 at 9:08
Aditya GoyalAditya Goyal
11
answered Dec 29 '18 at 9:08
Aditya GoyalAditya Goyal
11
answered Dec 29 '18 at 9:08
Aditya GoyalAditya Goyal
11
11
thats possible but more memory hungry and slower, if theres another better method to do it that would be preferred.
– Sharan Narasimhan
Dec 29 '18 at 9:12
add a comment |
thats possible but more memory hungry and slower, if theres another better method to do it that would be preferred.
– Sharan Narasimhan
Dec 29 '18 at 9:12
thats possible but more memory hungry and slower, if theres another better method to do it that would be preferred.
– Sharan Narasimhan
Dec 29 '18 at 9:12
thats possible but more memory hungry and slower, if theres another better method to do it that would be preferred.
– Sharan Narasimhan
Dec 29 '18 at 9:12
add a comment |
Using Numpy and your example:
import numpy as np
l = [1,3,5,6,8,10]
l2 = np.reshape(l,[2,-1])
>>> l2
array([[ 1, 3, 5],[ 6, 8, 10]])
It looks like you're working with sound data so I would highly recommend using the numpy module as vectorizing array operations will be far faster than using a for loop with Python objects (up to 100x faster in some cases).
The Numpy module was designed specifically for these type of use cases so I would encourage you to learn how to use it.
answered Dec 29 '18 at 9:03
Jason BaumgartnerJason Baumgartner
292110
Yes okay I will consider using numpy, however its tough to use reshape all the time because all sound samples do not have the same size. But thanks I can probably work around this problem with a little bit more code
– Sharan Narasimhan
Dec 29 '18 at 9:16
np.reshape(l,[2,-1]) -- the -1 here means to auto-adjust to the size so as long as the array is even sized, it will work. If you could give more info on the structure of the data, I might be able to give you some better ideas.
– Jason Baumgartner
Dec 29 '18 at 9:21
add a comment |
Using Numpy and your example:
import numpy as np
l = [1,3,5,6,8,10]
l2 = np.reshape(l,[2,-1])
>>> l2
array([[ 1, 3, 5],[ 6, 8, 10]])
It looks like you're working with sound data so I would highly recommend using the numpy module as vectorizing array operations will be far faster than using a for loop with Python objects (up to 100x faster in some cases).
The Numpy module was designed specifically for these type of use cases so I would encourage you to learn how to use it.
answered Dec 29 '18 at 9:03
Jason BaumgartnerJason Baumgartner
292110
Yes okay I will consider using numpy, however its tough to use reshape all the time because all sound samples do not have the same size. But thanks I can probably work around this problem with a little bit more code
– Sharan Narasimhan
Dec 29 '18 at 9:16
np.reshape(l,[2,-1]) -- the -1 here means to auto-adjust to the size so as long as the array is even sized, it will work. If you could give more info on the structure of the data, I might be able to give you some better ideas.
– Jason Baumgartner
Dec 29 '18 at 9:21
add a comment |
Using Numpy and your example:
import numpy as np
l = [1,3,5,6,8,10]
l2 = np.reshape(l,[2,-1])
>>> l2
array([[ 1, 3, 5],[ 6, 8, 10]])
It looks like you're working with sound data so I would highly recommend using the numpy module as vectorizing array operations will be far faster than using a for loop with Python objects (up to 100x faster in some cases).
The Numpy module was designed specifically for these type of use cases so I would encourage you to learn how to use it.
answered Dec 29 '18 at 9:03
Jason BaumgartnerJason Baumgartner
292110
Using Numpy and your example:
import numpy as np
l = [1,3,5,6,8,10]
l2 = np.reshape(l,[2,-1])
>>> l2
array([[ 1, 3, 5],[ 6, 8, 10]])
It looks like you're working with sound data so I would highly recommend using the numpy module as vectorizing array operations will be far faster than using a for loop with Python objects (up to 100x faster in some cases).
The Numpy module was designed specifically for these type of use cases so I would encourage you to learn how to use it.
answered Dec 29 '18 at 9:03
Jason BaumgartnerJason Baumgartner
292110
edited Dec 29 '18 at 9:09
answered Dec 29 '18 at 9:03
Jason BaumgartnerJason Baumgartner
292110
answered Dec 29 '18 at 9:03
Jason BaumgartnerJason Baumgartner
292110
answered Dec 29 '18 at 9:03
Jason BaumgartnerJason Baumgartner
292110
292110
Yes okay I will consider using numpy, however its tough to use reshape all the time because all sound samples do not have the same size. But thanks I can probably work around this problem with a little bit more code
– Sharan Narasimhan
Dec 29 '18 at 9:16
np.reshape(l,[2,-1]) -- the -1 here means to auto-adjust to the size so as long as the array is even sized, it will work. If you could give more info on the structure of the data, I might be able to give you some better ideas.
– Jason Baumgartner
Dec 29 '18 at 9:21
add a comment |
Yes okay I will consider using numpy, however its tough to use reshape all the time because all sound samples do not have the same size. But thanks I can probably work around this problem with a little bit more code
– Sharan Narasimhan
Dec 29 '18 at 9:16
np.reshape(l,[2,-1]) -- the -1 here means to auto-adjust to the size so as long as the array is even sized, it will work. If you could give more info on the structure of the data, I might be able to give you some better ideas.
– Jason Baumgartner
Dec 29 '18 at 9:21
Yes okay I will consider using numpy, however its tough to use reshape all the time because all sound samples do not have the same size. But thanks I can probably work around this problem with a little bit more code
– Sharan Narasimhan
Dec 29 '18 at 9:16
Yes okay I will consider using numpy, however its tough to use reshape all the time because all sound samples do not have the same size. But thanks I can probably work around this problem with a little bit more code
– Sharan Narasimhan
Dec 29 '18 at 9:16
np.reshape(l,[2,-1]) -- the -1 here means to auto-adjust to the size so as long as the array is even sized, it will work. If you could give more info on the structure of the data, I might be able to give you some better ideas.
– Jason Baumgartner
Dec 29 '18 at 9:21
np.reshape(l,[2,-1]) -- the -1 here means to auto-adjust to the size so as long as the array is even sized, it will work. If you could give more info on the structure of the data, I might be able to give you some better ideas.
– Jason Baumgartner
Dec 29 '18 at 9:21
add a comment |
Assuming that the array soundData is a 7 X 22000 holds your data. So creating a new array cleanSoundData of size 7 x 220 can be done as follows. Even it is more generalized whether its 2 x 1000 or 1000 x 50000.
cleanSoundData =
for i in range(len(soundData)):
cleanSoundData.append() # adding new row
for j in range(0, len(soundData[i]), 100):
cleanSoundData[i].append(soundData[i][j]) # adding data to the row
Hope this will work for you.
answered Dec 29 '18 at 9:26
Anidhya BhatnagarAnidhya Bhatnagar
46329
add a comment |
Assuming that the array soundData is a 7 X 22000 holds your data. So creating a new array cleanSoundData of size 7 x 220 can be done as follows. Even it is more generalized whether its 2 x 1000 or 1000 x 50000.
cleanSoundData =
for i in range(len(soundData)):
cleanSoundData.append() # adding new row
for j in range(0, len(soundData[i]), 100):
cleanSoundData[i].append(soundData[i][j]) # adding data to the row
Hope this will work for you.
answered Dec 29 '18 at 9:26
Anidhya BhatnagarAnidhya Bhatnagar
46329
add a comment |
Assuming that the array soundData is a 7 X 22000 holds your data. So creating a new array cleanSoundData of size 7 x 220 can be done as follows. Even it is more generalized whether its 2 x 1000 or 1000 x 50000.
cleanSoundData =
for i in range(len(soundData)):
cleanSoundData.append() # adding new row
for j in range(0, len(soundData[i]), 100):
cleanSoundData[i].append(soundData[i][j]) # adding data to the row
Hope this will work for you.
answered Dec 29 '18 at 9:26
Anidhya BhatnagarAnidhya Bhatnagar
46329
Assuming that the array soundData is a 7 X 22000 holds your data. So creating a new array cleanSoundData of size 7 x 220 can be done as follows. Even it is more generalized whether its 2 x 1000 or 1000 x 50000.
cleanSoundData =
for i in range(len(soundData)):
cleanSoundData.append() # adding new row
for j in range(0, len(soundData[i]), 100):
cleanSoundData[i].append(soundData[i][j]) # adding data to the row
Hope this will work for you.
answered Dec 29 '18 at 9:26
Anidhya BhatnagarAnidhya Bhatnagar
46329
answered Dec 29 '18 at 9:26
Anidhya BhatnagarAnidhya Bhatnagar
46329
answered Dec 29 '18 at 9:26
Anidhya BhatnagarAnidhya Bhatnagar
46329
answered Dec 29 '18 at 9:26
Anidhya BhatnagarAnidhya Bhatnagar
46329
46329
add a comment |
add a comment |
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Is using the numpy module an option for you?
– Jason Baumgartner
Dec 29 '18 at 8:58